Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises: 53

Answer

$$y=-x+\pi$$

Work Step by Step

Finding the derivative of the curve: $y'=\frac{d\sin(\sin{x})}{dx}$ Using the chain rule: $y'=\frac{d\sin(\sin{x})}{d\sin{x}} \times \frac{d\sin{x}}{dx}$ $=\cos(\sin{x}) \times\cos{x}$ Finding the slope ($m$) to the tangent line at ($\pi$,0): $m=y'(\pi)$ $=\cos(\sin{\pi}) \times\cos{\pi}=\cos(0)\times(-1)=1\times(-1)=-1$ From the point-slope form of a linear equation: $y-0=-1(x-\pi)$ Thus, an equation of the tangent line would be: $y=-x+\pi$
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